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olasank [31]
2 years ago
5

What is the picture for ?

Chemistry
1 answer:
ASHA 777 [7]2 years ago
6 0

Answer:concentration

Explanation:

In the first, the particles are much more concentrated, or close together, than in the second, where the particles are much more sparse.

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What minimum volume of 0.200 m potassium iodide solution is required to completely precipitate all of the lead in 195.0 ml of a
lbvjy [14]
First, we write the reaction equation:

2KI + PbNO₃  → K₂NO₃ + PbI₂
The molar ratio of KI to PbNO₃ is 2 : 1
Moles of PbNO₃ present:
Moles = concentration (M) x volume (dm³)
= 0.194 x 0.195
= 0.038
Moles of KI required = 2 x 0.038 = 0.076 moles
concentration = moles / volume
volume = moles / concentration
= 0.076 / 0.2
= 0.38 L = 380 ml
7 0
3 years ago
For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
Diano4ka-milaya [45]
The balanced chemical reaction is given as follows:

<span>2 KClO3(s) → 2 KCl(s) + 3 O2(g)

The starting amount of the reactant are given above. These values would be used for the calculations. We do as follows:

</span>2.72 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 1.06 g O2
<span>
0.361 g KClO3 </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.14 g O2
<span>
83.6 kg KClO3 (1000g / 1kg) </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2
<span>
22.5 mg KClO3</span> (1 g / 1000 mg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.009 g O2
4 0
2 years ago
Read 2 more answers
A solution is known to contain only one type of cation. Addition of Cl1- ion to the solution had no apparent effect, but additio
zhannawk [14.2K]

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate SO_4~^-^2 and the chloride Cl^-:

<u>Sulfate</u>

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.(Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2), which are NOT soluble.

<u>Chloride</u>

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). (Pb^+^2~,Ag^+~,Hg_2~^+^2), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I),  lead" and our possibilities are:

<u>"Calcium, strontium, or barium".</u>

I hope it helps!

7 0
3 years ago
What is responsible for the larger size of an anion in comparison with the atom from which it is form?
GarryVolchara [31]

Answer:

electron-electron repulsion

Explanation:

When electrons add into valence shell of neutral elements, the element assumes a negative oxidation state. With this, the number of electrons having (-) charges will be larger than the number of protons having positive (+) charges. As a result, the extra electrons repel one another (i.e., like charges repel) and a larger radius is the result.  

In contrast, when cations are formed, electrons are removed from the valence level (oxidation) producing an element having a greater number of protons than electrons. The larger number of protons will function to attract the electron cloud with a greater force that results in a contraction of atomic radius and a smaller spherical volume than the neutral unionized element.    

To visualize, see attached chart that shows atomic and ionic radii before and after ionization of the elements.  

Download pdf
4 0
3 years ago
What is the molar mass of C4H10
Phantasy [73]
Your answer would be 58.12g/mol ;)
7 0
3 years ago
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